# The oxidation number of nitrogen in the following compounds.

1. Oct 5, 2006

### alancj

I worked on this problem for quite a while last night and I'm not at all sure if my answers are correct. This is the problem I’ve been given:

I know nitrogen exists in the 3-, 3+ and 5+ oxidation states. I would think that they would give me at least one of each. I'm REALLY not sure about b, c, and d.

For b. I don't really know how to handle the 1- charge. Since the N in NO2 is 3+ normally, I figured that maybe the extra charge makes it 5+ (since there is no 4+ state).
For c. my first thought was "for which nitrogen?" I would think one N would be 3+ and the other 3-. Should I give my answer for both?
For d. I reasoned that it was similar to b. except that instead of being a 1- ion it has a chloride attached, so I gave it the same oxidation # (5+) as b.

I’m more confident that I have a. and e. right though I wouldn’t bet my life on it.

My chemistry book doesn't really seem to address this very much. At least not in the form of this problem... If it does I must be missing it. I have read every one of the first 375 pages in the book, 14 chapters, and I am now on my mid-term exam for those chapters. So this info should be in the pages I've already read.

So, does anyone know what I’m getting right or wrong? Some help for any that I got wrong would be great.

Thanks,
Alan

Last edited: Oct 5, 2006
2. Oct 5, 2006

### Stevedye56

A is wrong because each H atom is +1 you have 3 so its +3 but the over all charge is 0. 3+x=0. B is correct. C is incorrect becasue atoms in their elemental state are always 0. D is correct. E is incorrect. Each H atom is +1 so you have +4, overall charge of 0. 4+x=0

Last edited: Oct 5, 2006
3. Oct 5, 2006

### alancj

So... For a. N has an oxidation of 3-? I guess I was thinking of the 1- hydride ion giving N a 3+, how would I know which one in the future?

With e. are you saying it should be -4? Because I thought that wouldn't be possible (only 3-, 3+, 5+). I figured it was 3+ because the oxidation # for N in the N2F4 is 3+ (according to www.webelements.com) and that N2H4 would be the same oxidation number. Unless the hydrogen is 1+ instead of 1- then it could be 3-.

Alan

4. Oct 5, 2006

### Stevedye56

"So... For a. N has an oxidation of 3-? I guess I was thinking of the 1- hydride ion giving N a 3+, how would I know which one in the future?"

Are you talking about the element by itself here? ex H_2 or Na or C.

Yes for e i am saying it is -4 because oxygen is definate and nitrogen will "morph" to what it needs to be to get the appropriate overall charge.

Ill try to get you a link if this is confusing

http://wine1.sb.fsu.edu/chm1045/notes/Bonding/OxNum/Bond10.htm

Last edited by a moderator: Apr 22, 2017
5. Oct 5, 2006

### cheechnchong

a) +3 (because 3 Hydrogens with +1 charge = +3)
b) +3 (because O = 2-(2)-- there's two oxygens--and the overall charge is -1; therefore nitrogen needs to be +3 to match overall charge of -1)
c) 0 (because any element alone has a 0 charge...no matter what)
d) 5+ (because again -2(2)= -4 Oxygen and Cl = -1...add those up and you get -5. So you need 5+ Nitrogen for 0 charge)
e) -2 (because 4 H = +4 and you already have 2 N...so the number above--oxidation-- is -2 to give an overall 0 charge)

6. Oct 5, 2006

### 3trQN

According to the Periodic table (if you have a good one), N takes the following oxidation states: (+/-3),5,4,2

a) is +3

Hydrogen has the following oxidation styates ,+1 or -1 ( donor or acceptor ).

b) NO_2 is (IV) or +3, Oxygen is -2 and there are 2 bonded with N. Here oxidation is obvious, N is Oxidised by O. The overall charge on the ion shows that it has gain some charge from somewhere, so it is added to the mix ( N has to give up one less since its been aquired elsewhere ).

c) 0, the overall charge is 0, but thats not the question. Again the oxidation number reflects the extra charge aquired/lost, and since N2 is a triple bond, both N atoms are both donating and accepting the same number of electrons and so are 0 oxidation number.

OILRIG - Oxidation is loss Reduction is gain

Oxidation number represents the number of electrons that an atom is donating to form a bond. If its acting as an acceptor (which largley depends on the electronegativity of the bonding species) then its Oxidation number will be negative.

d)NO2Cl - Not come accross this compound much, following the rules for bonding it would be +5. Since thats a valid oxidation state for N it looks cool.

e)N2H4 - Hydrazine (rocket Fuel) - Oxidation state +2, Each H pair is -2 and the N-N bond is non polar (equal donation/acceptance of e-).

The concept of electron ownership is silly, but its just a matter of keeping track of the total quantity of charge balencing things up wrt the conservation of charge and charge distribution in a molecule/compound.

Last edited: Oct 6, 2006
7. Oct 5, 2006

### alancj

Thanks for the link. I guess my + and - signs should be in front of the numbers for charge.

And yes I'm talking about the oxidation state of N for a.

In a. if the 3 hydrogen’s are +1 then N would have to be -3 to make it equal to 0... Makes sense.

But cheechnchong and 3trQN, why do you say a. should be +3?

-Alan

Last edited by a moderator: Apr 22, 2017
8. Oct 5, 2006

### cheechnchong

hahha OUCH! but i think (a) should be -3 ... sorry! because +3 = H3 and Nitrogen needs -3 for overall charge of 0

Last edited: Oct 5, 2006
9. Oct 5, 2006

### alancj

I have a few more questions... for e. -2 makes sense but how is that "allowed" given that there is only +/- in front of the 3 but not the 2 in the above list of oxidation states?

Also, I was looking at this page: http://www.webelements.com/webelements/compounds/text/N/N1O2-10102440.html" and the name for NO2 is nitrogen (IV) oxide, and the oxidation state for N is said to be +3. Firstly, why does the name have a IV (4) in it instead of III (3)? Where does the IV come from? Secondly, in the case of b., the NO2 is an ion with 1- charge, so how could N have the same oxidation state as when NO2 had a charge of zero?

Thanks,
Alan

Edit: Stevedye56 said that B. was correct (+5). So if that was the case then 5 + (-4)=1 and if the 1 just means an extra electron (which has a -1 charge) then that should make the whole ion 1-.

Come to think of it, shouldn’t the N in NO2 (with no charge) have an oxidation state of +4? Since there are 2O with a -2 each then 4+(-4)=0. Which would make the webelements.com page I was looking at wrong?

Last edited by a moderator: Apr 22, 2017
10. Oct 6, 2006

### Stevedye56

Im very sorry for that misread i did not notice that negative sign because im used to them being in front. This means that it should be +3, Im very sorry i didnt notice that and I hope i didnt confuse you

Last edited by a moderator: Apr 22, 2017
11. Oct 6, 2006

### cheechnchong

Hey, i just looked at the webelements.com page...the O (oxygen) is always -2 (unless it's a peroxide like H2O2, where O is -1)...now on the other hand, nitrogen is not a fixed charged (it just balances out charge in compound)

Last edited by a moderator: Apr 22, 2017
12. Oct 6, 2006

### alancj

Ok, but I don't see why the oxidation # for nitrogen is stated as +3 in the NO2 molecule (I’m not talking about my NO2 Ion here!) as webelements.com says. With oxygen being fixed at -2 as you say... that would mean, having 2 oxygen’s, that nitrogen would have to be +4 in order to give an overall zero charge. So I'm thinking that webelements.com page is wrong on this. Unless I'm completely missing something.

I'm still not sure why, with e. (N2H4) nitrogen could have -2, since nitrogen only takes (+/-3),5,4,2. (unless I missing something again) The math makes sense and if there was a +/- thingy in front of the 2 then I'd feel better. Is it possible that the hydrogen is taking a charge of -1 giving a total of -4 instead of +4? Thus giving nitrogen a "legal" charge of +2 each?

-Alan

13. Oct 6, 2006

### cheechnchong

all i can tell you here is that the charge of Nitrogen is not a fixed value like Fluorine, Oxygen, Groups 1,2,6,7 Elements, and a few more (i think from the transition metals). However, the strategy in approaching oxidation problems is to assign charges to the ones you know about (i.e. Na, Ca, Cl, O, I, F, etc etc)...and then assign charges to elements like Nitrogen in order to balance out to an overall charge of 0 (or if a compound is ionized to charges of -1,-2, etc--overall charge should keep these values). Makes sense?

14. Oct 6, 2006

### 3trQN

Its (IV) and thats +4, the +3 must be a typo. The (IV) is the oxidation number in numerals, as seen with Fe(II) etc.

The question now is, if the formal charge on the molecule adds to or subtracts from the oxidation number of N. ( With respect to question (b) where the $$[NO_{2}]^{-}$$ has a neg charge)

Well i think its logical that if there is an overall net gain in charge on the molecule that ( this whole idea of electron ownership is just silly, but ill keep it up even though it irritates me more each time i use it ) there is one less electron that N has to "loose" i.e. oixidised less...therefor a lower oxidation number....therefor +3.

Im by no means certain, so I refer to a textbook:

$$(+3)+2(-2) = -1$$ QED wrt (b)

Last edited by a moderator: May 2, 2017
15. Oct 6, 2006

### alancj

Ok, so are saying that nitrogen can have whatever oxidation # (+/-) that you can write?

From what I have found on the internet hydrogen would take a +1 state when it bonds with nonmetals, and a -1 state when it bonds with metals. Also I found on a site this rule "In Binary Compounds, the more Electronegative (EN) element is assigned to have a negative oxidation number." Nitrogen is 3.07 and hydrogen is 2.20 so that must mean that in the case of problem e. hydrogen would have an oxidation # of +1 and that my nitrogen will have to be -2. I guess you guys are right then!

SO… in summary, webelements.com has a typo that messed me up, Nitrogen can have any oxidation state possible +/- (2,3,4,5), and it helps to have all the rules from the start.

A: -3
B: +3
C: 0
D: +5
E: -2

Thanks, for all the help guys
Alan

16. Oct 6, 2006

### cheechnchong

it can have whatever as long as it contributes to the overall charge of 0 (or if it's an ion -1, -2, +1, etc)